theorem Th36:
  o in Day(No_Ord A,B) & B c= A implies o in Day B
proof
  assume A1:o in Day(No_Ord A,B) & B c= A;
  then No_Ord A /\ [:BeforeGames B,BeforeGames B:] =
  No_Ord B /\ [:BeforeGames B,BeforeGames B:] by Th31;
  hence thesis by A1,Th10;
end;
